How do you solve Number of Changing Keys on LeetCode?

Number of Changing Keys (LeetCode #3019) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Changing keys is only counted when the letter changes, not the case.

What is the brute force solution for Number of Changing Keys?

The brute force approach for Number of Changing Keys is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every character in the string against the previous character to see if a key change occurred. This is simple but inefficient as it may involve unnecessary comparisons. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Changing Keys?

Number of Changing Keys uses the following concepts: String. The recommended patterns to study are: String Manipulation, Two Pointers.

What are the interview tips for Number of Changing Keys?

Explain your thought process clearly while coding. Consider edge cases like strings with only one character. Discuss the time and space complexities of your solution.

What are common mistakes when solving Number of Changing Keys?

Not considering case sensitivity properly. Failing to initialize the counter correctly.

#3019

Number of Changing Keys

Easy

StringString ManipulationTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution simplifies the problem by converting all characters to lowercase and then counting changes. This reduces the complexity by eliminating the need for case comparisons.

⚙️

Algorithm

4 steps

1Step 1: Convert the entire string to lowercase.
2Step 2: Initialize a counter to zero for key changes.
3Step 3: Loop through the string from the second character to the end.
4Step 4: Compare each character with the previous one. If they are different, increment the counter.

solution.py12 lines

1# Full working Python code
2
3def count_key_changes(s):
4    s = s.lower()
5    changes = 0
6    for i in range(1, len(s)):
7        if s[i] != s[i - 1]:
8            changes += 1
9    return changes
10
11# Example usage
12print(count_key_changes('aAbBcC'))  # Output: 2

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Complexity note: The time complexity is O(n) because we traverse the string once. The space complexity is O(n) due to the creation of a new string when converting to lowercase.

1Changing keys is only counted when the letter changes, not the case.
2Converting to lowercase simplifies the comparison process.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.